Quantitative Aptitude Questions Daily Quiz Day – 22

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Dear Aspirants, Our SSC Crackers team is providing a new series of Quantitative Aptitude Questions for Upcoming Exam so the aspirants can practice it on a daily basis. These questions are framed by our skilled experts after understanding your needs thoroughly. Aspirants can practice these new series questions daily to familiarize with the exact exam pattern and make your preparation effective.

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1) The length of a string between a kite and a point on the ground is 111. If the string makes an angle θ with the level ground such that tanθ = 35/12, then how high is the kite?

a) 101 m

b) 105 m

c) 95 m

d) 108 m

2) The radius and height of a cylindrical vessel are in the ratio of 2:5, respectively. If volume of the vessel is 1620 cm3, then find the curved surface area of the vessel. (Take π = 3)

a) 560 cm2

b) 580 cm2

c) 510 cm2

d) 540 cm2

3) In the given triangle, BE and AD are two medians, ∠AFB = 90o, the length of AD = 18 cm and the length of BE = 12 cm, then what is the length of BC?

a) 28 cm

b) 20 cm

c) 16 cm

d) 24 cm

4) In the figure given below AB || CD and lines PQ and ML are two traversal line, then what is the value of ‘x’?

a) 750

b) 650

c) 800

d) 700

5) In the figure given below O is the centre of the circle, ∠ABO = 250 and ∠ACO = 150, then what is the measure of ∠BAC?

a) 400

b) 500

c) 600

d) 700

6) A train of length 240 m crosses a tree in 20 seconds. Find the time taken by the same train to cross a bridge of length 480 m.

a) 60 seconds

b) 65 seconds

c) 54 seconds

d) 62 seconds

7) 3 litres of 40% spirit solution is mixed with 4 litres of x% spirit solution to get a resultant mixture of 60% spirit. What is the value of x?

a) 65

b) 70

c) 75

d) 60

8) Find the average of the sum of prime numbers between 39 and 68.

a) 53

b) 57

c) 55

d) 45.2

9) A shopkeeper gives 20% discount on an article and makes a profit of 10%. If the cost price of the article is Rs 400 then what is the marked price?

a) Rs 400

b) Rs 600

c) Rs 500

d) Rs 550

10) At what rate of simple interest a sum of Rs. 5650 will yield an interest of Rs.1017 in 4 years?

a) 6%

b) 3.5%

c) 8.25%

d) 4.5%

Answers :

1) Answer: B)

Let AB = 35x and BC = 12x

(AC)2 = (AB)2 + (BC)2

1112 = 1225x2 + 144x2

x2 = 9

x = 3

Therefore, height of the kite = 35 x 3 = 105 m

2) Answer: D)

Let, radius and height of the cylindrical vessel be 2x cm and 5x cm respectively.

So, 3 × 2x × 2x × 5x = 1620

60x2 = 1620

x3 = 27, x = 3

So, radius and height of vessel are 6 cm and 15 cm, respectively.

So, curved surface area = 2 × 3 × 6 × 15 = 540 cm2

3) Answer: B)

Given that,

AD = 18 cm and BE = 12 cm

AD and BE are medians of the triangle

AF: FD = BF: FE = 2: 1

FD = 18/3 = 6 cm and BF = 12 * (2/3) = 8cm

Also given that,

∠AFB = 90o

So, ∠BFD = 90o

BD = (BF2 + FD2)1/2 = (64 + 36)1/2 = 10 cm

BC = 2 * BD = 20 cm

4) Answer: C)

∠AOM = 450 [Corresponding angles.]

∠LOB = ∠AOM = 450 [Vertically opposite angles.]

∠POB = 1250 [Corresponding angles.]

Now,

∠POB = ∠POL + ∠LOB = 1250

x0 + 45 = 125

x = 800

5) Answer: A)

In ∆BOC, OB = OC [Radius of the circle.]

Let ∠OBC = ∠OCB = x

∠BOC = 180 – (x + x) = (180 – 2x)

∠BAC = (180 – 2x)/2 = (90 – x) [In a circle the angle at the centre is twice the angle at the circumference.]

∆ABC,

∠ABC + ∠ACC + ∠BAC = 1800

(25 + x) + (90 – x) + (15 + x) = 180

x = 50

∠BAC = (90 – x) = 400

6) Answer: A)

Speed of train = 240/20 = 12 m/s

Required time taken = (240 + 480)/12 = 60 seconds

7) Answer: C)

By Alligation Method:

Given, (x – 60)/20 = 3/4

=> 4x – 240 = 60

=> 4x = 300

=> x = 75

8) Answer: A)

Prime numbers between 39 and 62 are 41, 43, 47, 53, 59, 61, 67

Required average = (41 + 43 + 47 + 53 + 59 + 61 + 67)/7 = 53

9) Answer: D)

Let the marked price be Rs 100a.

He gives a discount of 20% hence the price at which he sold the article

= 0.8 X 100a = 80a

Even after selling the article at 80a he is earning 10% profit.

Hence cost Price =Rs 80a/1.1

According to the question cost Price = Rs 400

Hence, 80a/1.1 = 400

=> a = (400 X 1.1)/80 = 5.5

So marked Price = 100a = 100 X 5.5 = Rs 550

10) Answer: D)

Let the rate of simple interest be ‘x’% p.a.

1017 = (5650 × x × 4)/100

x = 1017/226

x = 4.5%

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